Problem: Let $y=\dfrac{e^x}{x}$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{e^x-1}{x^2}$ (Choice B) B $\dfrac{e^x(x-1)}{x^2}$ (Choice C) C $\dfrac{e^{x-1}}{x^2}$ (Choice D) D $e^x$
Solution: $\dfrac{e^x}{x}$ is the quotient of two, more basic, expressions: $e^x$ and $x$. Therefore, $\dfrac{dy}{dx}$ can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left(\dfrac{e^x}{x}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(e^x)x-e^x\dfrac{d}{dx}(x)}{(x)^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{e^x\cdot x-e^x\cdot 1}{(x)^2}&&\gray{\text{Differentiate }e^x\text{ and }x} \\\\ &=\dfrac{e^x(x-1)}{x^2}&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{e^x(x-1)}{x^2}$